This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Initially, Q = √ 4 = 2 ܥ E = ∫ ܸ݀ݍ = ∫ ݍ ଶ ଶ dq = 8 / 3 E total = 8/3 – ф When t > ∞, Q = Q 1 + Q 2 2 = √ܸ + (1 x V) V = 1 ܧ = ∫ ݍ ଶ ଵ dq + 0.5 x C x V 2 E total, ∞ = 1/3 + 5/2 = 5/6 E discharged = E total , 0 – E total, ∞ = 8/3 – 5/6 = 1.83 J 3) Find the amplitudes of the sinusoids at Vo1 and Vo2 when vi is applied The emitter current I E is fixed at 1 mA. So whatever be the change in the base voltage, an ideal current source keeps the current in its path constant. Hence the small signal change in emitter current i e = 0 Therefore the small signal base current i b = i c / ( β + 1) = 0. Hence v Hence v be = 0 V. Here v b = v i and v e = v o2 . Therefore v o2 = v i. Also collector current i c = α x i e = 0 . Hence v o1 =  i c x R = 0V...
View
Full Document
 Fall '10
 ABIDI
 Resistor, Electrical impedance, Etotal

Click to edit the document details